What is the sum of the interior angles of a polygon that has 9 sides?

Find x- and y-intercepts. Write ordered pairs representing the points where the line crosses the axes. 2x+3y=6

sweet-ann [11.9K]

Answer:

x-intercept: (3, 0)

y-intercept: (0, 2)

Step-by-step explanation:

For a function like:

y = f(x)

The x-intercept is the value of x when y = 0

the y-intercept is the value of y when x = 0

Also remember that an ordered pair is written as (x, y).

In this case we have the equation:

2*x + 3*y = 6

For the x-intercept, we just replace y by zero in the equation, then we get:

2*x + 3*0 = 6

Solving this for x, we get:

2*x = 6

x = 6/2 = 3

Then in this case, the ordered pair for the x-intercept is (3, 0)

For the y-intercept, we just need to replace x by zero in the equation:

2*0 + 3*y = 6

Solving this for y, we get:

3*y = 6

y = 6/3 = 2

Then the ordered pair for the y-intercept is (0, 2)

8 0

3 years ago

How do I solve this?

lakkis [162]

Answer:

$\large\boxed{4\sqrt{-81}+\sqrt{-25}=41i}$

Step-by-step explanation:

$i=\sqrt{-1}\\\\\sqrt{-81}=\sqrt{(81)(-1)}=\sqrt{81}\cdot\sqrt{-1}=9i\\\\\sqrt{-25}=\sqrt{(25)(-1)}=\sqrt{25}\cdot\sqrt{-1}=5i\\\\4\sqrt{-81}+\sqrt{-25}=4(9i)+5i=36i+5i=41i$

3 0

4 years ago

Determine the formula for the nth term of the sequence:<br>-2,1,7,25,79,...

rodikova [14]

A plausible guess might be that the sequence is formed by a degree-4* polynomial,

$x_n = a n^4 + b n^3 + c n^2 + d n + e$

From the given known values of the sequence, we have

$\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}$

Solving the system yields coefficients

$a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4$

so that the n-th term in the sequence might be

$\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}$

Then the next few terms in the sequence could very well be

$\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}$

It would be much easier to confirm this had the given sequence provided just one more term...

* Why degree-4? This rests on the assumption that the higher-order forward differences of $\{x_n\}$ eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of $\{x_n\}$ by $\Delta^{k}\{x_n\}$ . Then

• 1st-order differences:

$\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}$

• 2nd-order differences:

$\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}$

• 3rd-order differences:

$\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}$

• 4th-order differences:

$\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}$

From here I made the assumption that $\Delta^4\{x_n\}$ is the constant sequence {15, 15, 15, …}. This implies $\Delta^3\{x_n\}$ forms an arithmetic/linear sequence, which implies $\Delta^2\{x_n\}$ forms a quadratic sequence, and so on up $\{x_n\}$ forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.